- #1
Kreizhn
- 743
- 1
Homework Statement
Let m be the Lebesgue measure on [itex] \mathbb R^d [/itex], and define the open sets [itex] O_n = \{ x \in \mathbb R^d : d(x,E) < \frac1n \} [/itex] where
[tex] d(A,B) = \inf\{ |x-y| : x \in A, y \in B \} [/tex]
1) Find a closed and unbounded set E such that [itex] \lim_{n\to\infty} m(O_n) \neq m(E) [/itex].
2) Find an open and bounded set E such that [itex] \lim_{n\to\infty} m(O_n) \neq m(E) [/itex].
The Attempt at a Solution
This is technically the second part to question, which was to show that if E is compact then the limits do in fact hold. I think I might have the first part. In particular, I've just chosen a countable number of copies of the cantor set C in [itex] \mathbb R[/itex]. This set has zero measure, but [itex] m(O_n) = \infty [/itex] for every [itex] n \in \mathbb N [/itex] and so the limits don't converge.
My real issue is trying to find an open bounded set. My first thoughts were to try the complement to the cantor set [itex] [0,1]\setminus C [/itex] but this doesn't seem to produce the desired results. In particular, I know that we must have that [itex] m(E) < m(O_n) [/itex] by monotonicity, but since E is open (and we'll assume non-empty) then m(E) > 0, so there won't be any 0 = infinity mumbo jumbo here. In particular, it seems to me that the set I want will be one such that m(E) is not the same as m(cl E) where cl E is the closure of E. That is, the boundary of E will have non-zero measure. I just can't think of such a set.
Any help would be appreciated.